POJ 3280 Cheapest Palindrome(区间dp)

Cheapest Palindrome

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 2000+5;
int dp[maxn][maxn];
char str[maxn];
int cost[maxn];

int n, m;
int main() {
    //freopen("data.in", "r", stdin);
    while(scanf("%d %d", &n, &m) == 2) {
        scanf("%s", str);

        char c[2];
        int x, y;
        for(int i = 0; i < n; i++) {
            scanf("%s%d%d", c, &x, &y);
            cost[c[0] - 'a'] = min(x, y);
        }


        for(int i = 0; i < m; ++i)
            dp[i][i] = 0;

        for(int k = 2; k <= m; ++k) {//区间长度
            for(int i = 0; i < m - k + 1; i++) {
                int j = i + k - 1;
                dp[i][j] = 0x3f3f3f3f;
                int mi = min(dp[i+1][j] + cost[str[i]-'a'], dp[i][j-1] + cost[str[j]-'a']);

                dp[i][j] = min(mi, dp[i][j]);

                if(str[i] == str[j])
                    dp[i][j] = min(dp[i][j], dp[i+1][j-1]);
            }
        }

        printf("%d\n", dp[0][m-1]);

    }

    return 0;
}